2x y分之4x²-4xy y²÷(4x²-y²)
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原式=(2x+y)/(5xy-6xy)×(1/2)(x+2y)=(x+y)/(-xy)×(1/2)(x+y)=(-xy)/(2x+2y)=(-xy)/2(x+y)=(1/2)×(-xy)/(x+y)当
原式=[x-y(x-y)2-y(x+y)(x+y)(x-y)]•xyy-1=(1x-y-yx-y)•xyy-1=1-yx-y•xyy-1=-xyx-y.故答案是:-xyx-y.
原式=(x+2y)(x-2y)/(x+y)^2*2x(x+y)/x+2y=2x(x-2y)/(x+y)很高兴为您解答,【学习宝典】团队为您答题.请点击下面的【选为满意回答】按钮,
=-(x+2y)(x-2y)/(x+y)²×2x(x+y)/(x-2y)=-2x(x+2y)/(x+y)=-(2x²+4xy)/(x+y)
=(2X/3Y)*(3X/4Y)*(4/xy)=(2x*3x*4)/(3y*4y*xy)=2x/(y^3)
因为x-y=4xy所以x-2xy-y=2xy2x+3xy-2y=11xyx-2xy-y分之2x+3xy-2y=5.5
x平方+y平方+2xy分之x平方-4y平方除以x平方+xy分之2y+x=(x-2y)(x+2y)/(x+y)²·x(x+y)/(x+2y)=x(x-2y)/(x+y)
题目是√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2如果是:√x^3+X^2y+1/4xy+√(1/4x^3)-X^2y+xy^2=(3/2)√x^3+xy/4+xy^2=(3
原式=[(x+y)2(x-y)(x+y)+-4xy(x-y)(x+y)]×(x+3y)(x-3y)(x+3y)(x-y)=x-3yx+y,由已知得(3x-2y)(x+y)=0,因为x+y≠0,所以3x
x²+y²+2xy分之x²-4y²÷x²+xy分之2y+x=(x+2y)(x-2y)/(x+y)²×x(x+y)/(x+2y)=x(x-2y
产生4种配子xyxyyyx有一种y有两种xy1y2xy1xy2y1y2产生的配子如上y1、y2是一个意思所以分为一组xyxyyy=1221
[4(xy-1)^2-(xy+2)(2-xy)]除以4分之1xy=[4x^2乘以y^2-8xy+4-(4-x^2y^2)]除以4分之1xy=(4x^2乘以y^2-8xy+4-4+x^2y^2)除以4分
由已知条件可得:x-y=-4xy,(也就是在等式两边乘xy后整理,即可),代入所求的式子里分母为-4xy+7xy,即3xy;分子为-4xy-2xy,即-6xy,化简得-2
=xy(1-2y^2)=-0.119616
(1)(2)(x-1分之x+2)+(x-1分之-3)=2乘以x-1得x+2-2=2x-2x-2x=-2x=2(3)(x+2分之1)+(x分之1)=x²+2x分之32乘以x(x+2)得x+x+
[(xy-2)(-xy-2)-4(xy-1)^2]除以(-xy)=[-x²y²+4-4(x²y²-2xy+1)]÷(-xy)=(-x²y²+
[(xy+2)(xy-2)-2x的平方y的平方+4]÷(-xy)当x=10,y=25分之1=[x^2y^2-4-2x^2y^2+4]÷(-xy)=[-x^2y^2]÷(-xy)=xy=10*1/25=
(xy-x的平方)÷xy分之x-y=x(y-x)*xy/(x-y)=-x^2y(x的平方+x分之4x的平方-1)乘(1-2x分之x+1)÷x分之1=[x(x+1)分之(2x+1)(2x-1)]乘(1-